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what is x^dx equal to?

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  what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-09-29 03:54 -0700
    Re: what is x^dx equal to? Roland Franzius <roland.franzius@uos.de> - 2018-09-29 13:43 +0200
      Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-09-29 09:33 -0700
    Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-09-30 22:55 -0700
      Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-09-30 23:08 -0700
        Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-09-30 23:23 -0700
          Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-10-01 11:05 -0700
            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-01 22:34 -0700
              Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-10-01 23:20 -0700
                Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-02 00:36 -0700
                  Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-10-02 13:48 -0700
                    Re: what is x^dx equal to? Sergio <invalid@invalid.com> - 2018-10-02 19:34 -0500
                    Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-04 00:02 -0700
                      Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-10-04 11:03 -0700
                        Re: what is x^dx equal to? Sergio <invalid@invalid.com> - 2018-10-04 13:36 -0500
                          Re: what is x^dx equal to? 7777777 <jussir@yahoo.com> - 2018-10-04 11:52 -0700
                            Re: what is x^dx equal to? Sergio <invalid@invalid.com> - 2018-10-04 20:59 -0500
                          Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-04 22:26 -0700
                            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-05 00:24 -0700
                            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-05 00:24 -0700
                            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-05 00:24 -0700
                            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-05 00:24 -0700
                            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-05 00:24 -0700
                            Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-05 00:25 -0700
                        Re: what is x^dx equal to? Python <python@python.invalid> - 2018-10-04 21:39 +0200
    Re: what is x^dx equal to? Mathedman <Mathedman@hotmail.com> - 2018-10-02 09:10 -0500
    Re: what is x^dx equal to? Jan <filmart@gmail.com> - 2018-10-02 09:25 -0700
    Re: what is x^dx equal to? Sergio <invalid@invalid.com> - 2018-10-02 19:35 -0500
    Re: what is x^dx equal to? Mathedman <Mathedman@hotmail.com> - 2018-10-07 08:47 -0500
      Re: what is x^dx equal to? Zelos Malum <zelos.malum@gmail.com> - 2018-10-08 01:17 -0700

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#424509 — what is x^dx equal to?

What is x^dx equal to ?

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#424510

Am 29.09.2018 um 12:54 schrieb 7777777:
> What is x^dx equal to ?
> 

x^dx = 1+ log x dx

according to a^b = e^(b log a) and d^2=0

-- 

Roland Franzius

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#424521

lauantai 29. syyskuuta 2018 14.43.06 UTC+3 Roland Franzius kirjoitti:
> Am 29.09.2018 um 12:54 schrieb 7777777:
> > What is x^dx equal to ?
> > 
> 
> x^dx = 1+ log x dx
> 
> according to a^b = e^(b log a) and d^2=0
> 
> -- 
> 
> Roland Franzius

do you mean according to taking the Taylor series of a^b = e^(b log a) 
and dx^2 = 0 ?

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#424659

Den lördag 29 september 2018 kl. 12:55:02 UTC+2 skrev 7777777:
> What is x^dx equal to ?

define what you mean with dx here.

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#424662

maanantai 1. lokakuuta 2018 8.55.53 UTC+3 Zelos Malum kirjoitti:
> Den lördag 29 september 2018 kl. 12:55:02 UTC+2 skrev 7777777:
> > What is x^dx equal to ?
> 
> define what you mean with dx here.


dx is the same as in f'(x) = dy/dx

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#424665

Den måndag 1 oktober 2018 kl. 08:08:13 UTC+2 skrev 7777777:
> maanantai 1. lokakuuta 2018 8.55.53 UTC+3 Zelos Malum kirjoitti:
> > Den lördag 29 september 2018 kl. 12:55:02 UTC+2 skrev 7777777:
> > > What is x^dx equal to ?
> > 
> > define what you mean with dx here.
> 
> 
> dx is the same as in f'(x) = dy/dx

considering that is just notational convinience there, it still needs a proper definition to what it is.

neither dy nor dx in dy/dx refers to any number, x^dx is meaningless.
heck having dy and dx entirely free, in formal ways, is maeningless in most instances just because it literally refers to nothing but are part of a greater notation.

The choice of ntoation was however done by Leibniz and such just because that if you informally think of them as having a value and beign seperate, manipulation in calculus becomes very intuitive, while still being correct in the end. Even though that it is non-sensical foromally.

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#424717

maanantai 1. lokakuuta 2018 9.24.05 UTC+3 Zelos Malum kirjoitti:
> Den måndag 1 oktober 2018 kl. 08:08:13 UTC+2 skrev 7777777:
> > maanantai 1. lokakuuta 2018 8.55.53 UTC+3 Zelos Malum kirjoitti:
> > > Den lördag 29 september 2018 kl. 12:55:02 UTC+2 skrev 7777777:
> > > > What is x^dx equal to ?
> > > 
> > > define what you mean with dx here.
> > 
> > 
> > dx is the same as in f'(x) = dy/dx
> 
> considering that is just notational convinience there, it still needs a proper definition to what it is.
> 
> neither dy nor dx in dy/dx refers to any number, x^dx is meaningless.

no, it is not meaningless. 

The correct answer was already provided here:
https://groups.google.com/d/msg/sci.math/mEIGH8xomvc/yAmuBlz1AQAJ

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#424749

>no, it is not meaningless. 

You can give it meanign but if you say it is from dy/dx=f'(x), then it is meaningless cause dy nor dx does not refer to a function, value or anything on their own, but together they do.

He assumed you meant a value of some sort (at which it is correct) but you are not refering to a value with your previous statement. Therefore it is wrong.

you can defien dx^2=0, we got nilpotent structures and such, but it is not like dy/dx=f'(x) then.

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#424751

tiistai 2. lokakuuta 2018 8.34.41 UTC+3 Zelos Malum kirjoitti:
> >no, it is not meaningless. 
> 
> You can give it meanign but if you say it is from dy/dx=f'(x), then it is meaningless cause dy nor dx does not refer to a function, value or anything on their own, but together they do.
> 
> He assumed you meant a value of some sort (at which it is correct) but you are not refering to a value with your previous statement. Therefore it is wrong.
> 
> you can defien dx^2=0, we got nilpotent structures and such, but it is not like dy/dx=f'(x) then.

It is not wrong.

What function does dx refer in e^dx = 1 + dx ?

It comes from x^dx = 1 + ln x dx when x = e 

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#424757

Den tisdag 2 oktober 2018 kl. 08:20:42 UTC+2 skrev 7777777:
> tiistai 2. lokakuuta 2018 8.34.41 UTC+3 Zelos Malum kirjoitti:
> > >no, it is not meaningless. 
> > 
> > You can give it meanign but if you say it is from dy/dx=f'(x), then it is meaningless cause dy nor dx does not refer to a function, value or anything on their own, but together they do.
> > 
> > He assumed you meant a value of some sort (at which it is correct) but you are not refering to a value with your previous statement. Therefore it is wrong.
> > 
> > you can defien dx^2=0, we got nilpotent structures and such, but it is not like dy/dx=f'(x) then.
> 
> It is not wrong.
> 
> What function does dx refer in e^dx = 1 + dx ?
> 
> It comes from x^dx = 1 + ln x dx when x = e

The it is not the same as in dy/dx=f'

it is somethign else that you haven't specified, some real number presumibly

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#424795

tiistai 2. lokakuuta 2018 10.36.47 UTC+3 Zelos Malum kirjoitti:
> Den tisdag 2 oktober 2018 kl. 08:20:42 UTC+2 skrev 7777777:
> > tiistai 2. lokakuuta 2018 8.34.41 UTC+3 Zelos Malum kirjoitti:
> > > >no, it is not meaningless. 
> > > 
> > > You can give it meanign but if you say it is from dy/dx=f'(x), then it is meaningless cause dy nor dx does not refer to a function, value or anything on their own, but together they do.
> > > 
> > > He assumed you meant a value of some sort (at which it is correct) but you are not refering to a value with your previous statement. Therefore it is wrong.
> > > 
> > > you can defien dx^2=0, we got nilpotent structures and such, but it is not like dy/dx=f'(x) then.
> > 
> > It is not wrong.
> > 
> > What function does dx refer in e^dx = 1 + dx ?
> > 
> > It comes from x^dx = 1 + ln x dx when x = e
> 
> The it is not the same as in dy/dx=f'
> 
> it is somethign else that you haven't specified, some real number presumibly

y = f(x) = e^x 

f'(x) = dy/dx = (f(x+dx) - f(x))/dx =( e^(x+dx) - e^x )/dx = 
(e^x * e^dx - e^x)/dx = e^x(e^dx - 1)/dx = e^x((1+dx) - 1)/dx = e^x 

so e^dx = 1 + dx 
and 
dx is the same as in f'(x) = dy/dx

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#424811

On 10/2/2018 3:48 PM, 7777777 wrote:
> tiistai 2. lokakuuta 2018 10.36.47 UTC+3 Zelos Malum kirjoitti:
>> Den tisdag 2 oktober 2018 kl. 08:20:42 UTC+2 skrev 7777777:
>>> tiistai 2. lokakuuta 2018 8.34.41 UTC+3 Zelos Malum kirjoitti:
>>>>> no, it is not meaningless. 
>>>>
>>>> You can give it meanign but if you say it is from dy/dx=f'(x), then it is meaningless cause dy nor dx does not refer to a function, value or anything on their own, but together they do.
>>>>
>>>> He assumed you meant a value of some sort (at which it is correct) but you are not refering to a value with your previous statement. Therefore it is wrong.
>>>>
>>>> you can defien dx^2=0, we got nilpotent structures and such, but it is not like dy/dx=f'(x) then.
>>>
>>> It is not wrong.
>>>
>>> What function does dx refer in e^dx = 1 + dx ?
>>>
>>> It comes from x^dx = 1 + ln x dx when x = e
>>
>> The it is not the same as in dy/dx=f'
>>
>> it is somethign else that you haven't specified, some real number presumibly
> 
> y = f(x) = e^x 
> 
> f'(x) = dy/dx 

where f(x) = y

> = (f(x+dx) - f(x))/dx 

you left out the limit going to 0
f'(x) = limit as dx=>0 of ((f(x+dx) - f(x))/dx

> =( e^(x+dx) - e^x )/dx =

you left out the limit going to 0 again......

so sorry, https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus

> (e^x * e^dx - e^x)/dx = e^x(e^dx - 1)/dx = e^x((1+dx) - 1)/dx = e^x 
> 
> so e^dx = 1 + dx 
> and 
> dx is the same as in f'(x) = dy/dx
> 

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#424914

>> f'(x) = dy/dx = (f(x+dx) - f(x))/dx

FALSE!

It is

f'(x) = dy/dx = lim(h->0) (f(x+h) - f(x))/h

The limit must be there

and as others now have pointed out, your string of symbosl are meaningless cause you've not defined it properly and don't know basic calculus.

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#424933

torstai 4. lokakuuta 2018 10.03.05 UTC+3 Zelos Malum kirjoitti:
> >> f'(x) = dy/dx = (f(x+dx) - f(x))/dx
> 
> FALSE!
> 
> It is
> 
> f'(x) = dy/dx = lim(h->0) (f(x+h) - f(x))/h
> 
> The limit must be there
> 
> and as others now have pointed out, your string of symbosl are meaningless cause you've not defined it properly and don't know basic calculus.

you stupid jerk there is no limit in f'(x) = dy/dx

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#424936

On 10/4/2018 1:03 PM, 7777777 wrote:
> torstai 4. lokakuuta 2018 10.03.05 UTC+3 Zelos Malum kirjoitti:
>>>> f'(x) = dy/dx = (f(x+dx) - f(x))/dx
>>
>> FALSE!
>>
>> It is
>>
>> f'(x) = dy/dx = lim(h->0) (f(x+h) - f(x))/h
>>
>> The limit must be there
>>
>> and as others now have pointed out, your string of symbosl are meaningless cause you've not defined it properly and don't know basic calculus.
> 
> you stupid jerk there is no limit in f'(x) = dy/dx
> 

      you "dumbass"
 (far worser than "stupid jerk").



f'(x) = dy/dx   DOES NOT  = (f(x+dx) - f(x))/dx


you get an F for FAIL !


go post in   alt.math.for.babies

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#424941

torstai 4. lokakuuta 2018 21.37.00 UTC+3 Sergio kirjoitti:
> On 10/4/2018 1:03 PM, 7777777 wrote:
> > torstai 4. lokakuuta 2018 10.03.05 UTC+3 Zelos Malum kirjoitti:
> >>>> f'(x) = dy/dx = (f(x+dx) - f(x))/dx
> >>
> >> FALSE!
> >>
> >> It is
> >>
> >> f'(x) = dy/dx = lim(h->0) (f(x+h) - f(x))/h
> >>
> >> The limit must be there
> >>
> >> and as others now have pointed out, your string of symbosl are meaningless cause you've not defined it properly and don't know basic calculus.
> > 
> > you stupid jerk there is no limit in f'(x) = dy/dx
> > 
> 
>       you "dumbass"
>  (far worser than "stupid jerk").
> 
> 
> 
> f'(x) = dy/dx   DOES NOT  = (f(x+dx) - f(x))/dx
> 
> 
> you get an F for FAIL !
> 
> 
> go post in   alt.math.for.babies

fuck you

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#424968

On 10/4/2018 1:52 PM, 7777777 wrote:
> torstai 4. lokakuuta 2018 21.37.00 UTC+3 Sergio kirjoitti:
>> On 10/4/2018 1:03 PM, 7777777 wrote:
>>> torstai 4. lokakuuta 2018 10.03.05 UTC+3 Zelos Malum kirjoitti:
>>>>>> f'(x) = dy/dx = (f(x+dx) - f(x))/dx
>>>>
>>>> FALSE!
>>>>
>>>> It is
>>>>
>>>> f'(x) = dy/dx = lim(h->0) (f(x+h) - f(x))/h
>>>>
>>>> The limit must be there
>>>>
>>>> and as others now have pointed out, your string of symbosl are meaningless cause you've not defined it properly and don't know basic calculus.
>>>
>>> you stupid jerk there is no limit in f'(x) = dy/dx
>>>
>>
>>       you "dumbass"
>>  (far worser than "stupid jerk").
>>
>>
>>
>> f'(x) = dy/dx   DOES NOT  = (f(x+dx) - f(x))/dx
>>
>>
>> you get an F for FAIL !
>>
>>
>> go post in   alt.math.for.babies
> 
> fuck you
> 

no reason to get mad,
but you should not have called him a stupid jerk, be nice baby doodle.

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#424978

> you stupid jerk there is no limit in f'(x) = dy/dx 

It is by definition, look it up and nice going deleting everything!

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#424984

Den lördag 29 september 2018 kl. 12:55:02 UTC+2 skrev 7777777:
> What is x^dx equal to ?

As he is a coward I am reposting all

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#424985

Den lördag 29 september 2018 kl. 18:33:07 UTC+2 skrev 7777777:
> lauantai 29. syyskuuta 2018 14.43.06 UTC+3 Roland Franzius kirjoitti:
> > Am 29.09.2018 um 12:54 schrieb 7777777:
> > > What is x^dx equal to ?
> > > 
> > 
> > x^dx = 1+ log x dx
> > 
> > according to a^b = e^(b log a) and d^2=0
> > 
> > -- 
> > 
> > Roland Franzius
> 
> do you mean according to taking the Taylor series of a^b = e^(b log a) 
> and dx^2 = 0 ?

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